protons - Is there a $mathrm{p}^0$ particle?

In the following equation of a reaction

$$\mathrm{p}^0 + \mathrm{n} \to \mathrm{K}^+ + \Sigma^-$$

What is the quark composition of the $\mathrm{p}^0$ particle? Or is it supposed to be $\rho^0$?

For me it certainly looks more like $\mathrm p$, not $\mathrm \rho$.

(Source: K. A. Tsokos, Physics for the IB Diploma, Sixth Edition, Cambridge University Press)

Answer

It looks like a typo for $$\rm \rho^0 + n \to K^+ + \Sigma^-$$ where $\rho^0$ is the uncharged member of the isospin triplet with mass 770 MeV. According to the particle data group, the quark content of the light, unflavored mesons with isospin $I=1$ is $u\bar d, (u\bar u - d\bar d)/\sqrt2, d\bar u$.

You can tell that your "$p$" must be a meson, not a baryon, because both sides of the reaction must have the same baryon number, and the baryon number on the right side is $+1$.

Commenters on a duplicate question point out that it may also by a typo for $\pi^0$, if the printer's software produces the character "π" using the same code point as "p" but in some other typeface. Like the $\rho$, the $\pi$ also has zero baryon number and unit isospin; however the pion is spinless while the rho is a spin-one "vector" meson.